Distributions Integral Manifolds and the Frobenius Theorem

Assume $\mathcal{D}$ is a smooth distribution. By the local description of a [given] subbundle, each point has a neighborhood $U$ and a frame of $TM{\mathrm{\mid }}_U$ whose first $k$ members already frame $\mathcal{D}{\mathrm{\mid }}_U$. Those first $k$ sections are smooth vector fields, pointwise independent, and span the prescribed subspaces.

Conversely, assume such local vector fields exist near every point. On a [given] neighborhood $U$ where $X_1,\dots ,X_k$ are pointwise independent, their span is a rank-$k$ subbundle of $TM{\mathrm{\mid }}_U$, because in a local trivialization of $TM$ the columns formed by the $X_i$ have rank $k$ everywhere. Since that subbundle has fibres exactly ${\mathcal{D}}_q$, the family is a smooth distribution on $U$.

The two implications establish the equivalence. [given] ∎

If $Y,Z\in \mathrm{\Gamma }(\mathcal{D})$ and $f,g\in C^{\mathrm{\infty }}(M)$, then [given] $(fY+gZ{)}_p=f(p)Y_p+g(p)Z_p$ lies in the linear subspace ${\mathcal{D}}_p$ for every $p$. Hence $\mathrm{\Gamma }(\mathcal{D})$ is closed under addition and smooth scalar multiplication.

On $U$, the vectors $X_1(q),\dots ,X_k(q)$ form a basis of ${\mathcal{D}}_q$, [given] so each $Y_q\in {\mathcal{D}}_q$ has unique coefficients $f_i(q)$ with $Y_q={\sum }_i{f}_i(q)X_i(q)$. Because $Y$ and the frame fields are smooth, those coefficients are smooth on $U$.

Therefore the assignment $U\mapsto \mathrm{\Gamma }(\mathcal{D}{\mathrm{\mid }}_U)$ is locally free [given] of rank $k$: on every frame domain its section module is freely generated by that frame. This is a statement about the sheaf of local sections; it does not assert that the global module $\mathrm{\Gamma }(\mathcal{D})$ is free over $C^{\mathrm{\infty }}(M)$.

Relative to the dual coframe ${\omega }^1,\dots ,{\omega }^n$, the annihilator [given] bundle is locally spanned by ${\omega }^{k+1},\dots ,{\omega }^n$, because those and only those covectors vanish on the span of $X_1,\dots ,X_k$.

A tangent vector $v={\sum }_i{a}_i{X}_i$ is annihilated by every section of [given] ${\mathcal{D}}^{\circ }$ exactly when $a_{k+1}=\cdots =a_n=0$. Thus the double annihilator fibre is the span of $X_1,\dots ,X_k$, which is precisely $\mathcal{D}$.

Since the argument is pointwise and valid in every such neighborhood, [given] $({\mathcal{D}}^{\circ }{)}^{\circ }=\mathcal{D}$ as a subbundle of $TM$.

Because $i$ is an immersion, each $d{i}_q:T_{q}N\to T_{i(q)}M$ is injective. [given] Since $i$ is integral, its image is ${\mathcal{D}}_{i(q)}$, which has dimension $k$. Therefore $\dim T_{q}N=k$ for every $q\in N$.

The dimension of a manifold is the common dimension of its tangent spaces, [given] so $\dim N=k$.

Because $F{\mathrm{\mid }}_U$ is a diffeomorphism, its differential identifies [given] $TU\to TV$ fibrewise by linear isomorphisms. Transporting the rank-$k$ subbundle $\mathcal{D}{\mathrm{\mid }}_U$ through those isomorphisms yields a rank-$k$ smooth subbundle of $TV$, namely $F_{\ast }\mathcal{D}$.

The composite $F\circ i$ is an injective immersion, because both factors [given] are immersions and $F{\mathrm{\mid }}_U$ is injective. For each $q\in P$, $d(F\circ i{)}_q(T_{q}P)=d{F}_{i(q)}(d{i}_q(T_{q}P))=d{F}_{i(q)}({\mathcal{D}}_{i(q)})=(F_{\ast }\mathcal{D}{)}_{F(i(q))}\mathrm{.}$ Hence $F\circ i$ is an integral manifold of the transported distribution.

Therefore local diffeomorphisms preserve the regular-distribution and [given] integral-manifold structure on any neighborhood where they are genuine diffeomorphisms.

If $\mathcal{D}$ is involutive, then every bracket of tangent vector fields is tangent, so in particular every bracket $[X_i,X_j]$ is tangent on $U$.

Conversely, assume all frame brackets are tangent on $U$. Any tangent fields on $U$ have the form $X={\sum }_i{f}_i{X}_i$ and $Y={\sum }_j{g}_j{X}_j$ with smooth coefficients. Expanding $[X,Y]$ with the Leibniz rule expresses the bracket as a sum of terms involving the tangent fields $[X_i,X_j]$ and the frame fields $X_i$, hence again as a tangent field.

Since this holds on a neighborhood of every point, $\mathcal{D}$ is involutive exactly when one may check bracket closure on a local frame.

By integrability, the point $p$ lies on a connected integral manifold [given] $i:N\to M$ of $\mathcal{D}$ having the same dimension as the distribution. After shrinking near the point of $N$ over $p$, the immersion may be viewed as an embedding, so $X$ and $Y$ restrict to smooth vector fields $\tilde{X}$ and $\tilde{Y}$ on that local piece of $N$.

Along that local integral manifold, the fields $\tilde{X}$ and [given] $\tilde{Y}$ are $i$-related to $X$ and $Y$. Therefore their Lie bracket is $i$-related to $[X,Y]$. Since the bracket on $N$ is tangent to $N$, the value $[X,Y{]}_p$ lies in the image of $di$, which is ${\mathcal{D}}_p$.

The point $p$ and the tangent fields $X,Y$ were arbitrary, so [given] $[X,Y]\in \mathrm{\Gamma }(\mathcal{D})$. Hence $\mathcal{D}$ is involutive.

By the flow-box theorem there are local coordinates [given] $(t,u^2,\dots ,u^n)$ centered at $p$ in which $X_1={\mathrm{\partial }}_t$. Shrinking if necessary, write $X_j=a_j{\mathrm{\partial }}_t+{\sum }_{m=2}^n{b}_{jm}{\mathrm{\partial }}_{u^m}(2\le j\le k)$ and define $Z_j:=X_j-a_j{X}_1={\sum }_{m=2}^n{b}_{jm}{\mathrm{\partial }}_{u^m}\mathrm{.}$ Then each $Z_j$ is tangent to $\mathcal{D}$, has no ${\mathrm{\partial }}_t$-component, and $X_1,Z_2,\dots ,Z_k$ still form a local frame of $\mathcal{D}$.

Because $\mathcal{D}$ is involutive, each bracket $[X_1,Z_j]$ is tangent to [given] $\mathcal{D}$. It also has no ${\mathrm{\partial }}_t$-component, since $X_1={\mathrm{\partial }}_t$ and $Z_j$ has none. Therefore there are smooth functions $c_{j\mathrm{\ell }}$ such that $[X_1,Z_j]={\sum }_{\mathrm{\ell }=2}^k{c}_{j\mathrm{\ell }}{Z}_{\mathrm{\ell }}(2\le j\le k)\mathrm{.}$ For each fixed transverse coordinate, solve the matrix ODE ${\mathrm{\partial }}_{t}B=-C^{T}B,B(0,u)=I_{k-1},$ where $C=(c_{j\mathrm{\ell }})$. After shrinking again, the solution matrix $B$ is smooth and invertible.

For $j=2,\dots ,k$, set $Y_j:={\sum }_{\mathrm{\ell }=2}^k{B}_{\mathrm{\ell }j}{Z}_{\mathrm{\ell }}$. Because [given] the $Y_j$ are invertible linear combinations of $Z_2,\dots ,Z_k$, the fields $X_1,Y_2,\dots ,Y_k$ form a local frame of $\mathcal{D}$. Using the Leibniz rule for brackets with function coefficients and the differential equation for $B$, one gets $[X_1,Y_j]={\sum }_{\mathrm{\ell }=2}^k({\mathrm{\partial }}_t{B}_{\mathrm{\ell }j})Z_{\mathrm{\ell }}+{\sum }_{\mathrm{\ell }=2}^k{B}_{\mathrm{\ell }j}[X_1,Z_{\mathrm{\ell }}]=0.$ Hence each $Y_j$ commutes with $X_1$.

Therefore, after shrinking the neighborhood, there is a local frame [given] $X_1,Y_2,\dots ,Y_k$ of $\mathcal{D}$ with $[X_1,Y_j]=0$ for all $j\ge 2$.

Let ${\mathrm{\Phi }}^i$ be the local flow of $X_i$. Because the fields commute, their [given] local flows commute pairwise. Define $F(t_1,\dots ,t_k,s):={\mathrm{\Phi }}_{t_1}^1\circ \cdots \circ {\mathrm{\Phi }}_{t_k}^k(s)$ for $(t,s)$ near $(0,p)$ with $s\in S$. This map is smooth.

The differential of $F$ at $(0,p)$ sends the coordinate vector [given] ${\mathrm{\partial }}_{t_i}$ to $X_i(p)$ and the tangent space of $S$ identically into a complement of their span. Hence $d{F}_{(0,p)}$ is an isomorphism. By the inverse function theorem, after shrinking domains, $F$ is a local diffeomorphism.

In the resulting coordinates, changing only $t_i$ applies the $X_i$-flow, [given] so the pushforward of ${\mathrm{\partial }}_{t_i}$ is exactly $X_i$. Renaming the source coordinates as $(x^1,\dots ,x^n)$ gives the desired chart.

If $\mathcal{D}$ is integrable, then it is involutive by the necessity [given] proposition. This proves 1 => 2.

Now assume $\mathcal{D}$ is involutive. For $k=0$ the distribution is [given] zero, and for $k=n$ it is all of $TM$, so the displayed coordinate form is immediate. Thus only the case $1\le k<n$ needs work.

Choose a local frame $X_1,\dots ,X_k$ of $\mathcal{D}$ near $p$ with [given] $X_1(p)\mathrm{\ne }0$. By the frame-reduction lemma, after shrinking there are local sections $Y_2,\dots ,Y_k$ such that $X_1,Y_2,\dots ,Y_k$ frames $\mathcal{D}$, each $Y_j$ is tangent to the slices of a flow-box chart for $X_1$, and $[X_1,Y_j]=0$ for all $j\ge 2$. Let $S$ be the slice $x^1=0$ in that flow-box chart, and write $V_j:=Y_j{\mathrm{\mid }}_S$. Then $V_2,\dots ,V_k$ are pointwise independent vector fields on the $(n-1)$-manifold $S$. Because $[Y_i,Y_j]\in \mathrm{\Gamma }(\mathcal{D})$ and each $Y_j$ is tangent to the slices, the restrictions $[V_i,V_j]=[Y_i,Y_j]{\mathrm{\mid }}_S$ lie in the span of $V_2,\dots ,V_k$. Hence those $V_j$ span an involutive rank-$(k-1)$ distribution on $S$.

Apply the theorem inductively on the rank to that distribution on $S$. [given] There are local coordinates $(x^2,\dots ,x^k,x^{k+1},\dots ,x^n)$ on $S$ in which $\mathrm{span}(V_2,\dots ,V_k)=\mathrm{span}({\mathrm{\partial }}_{x^2},\dots ,{\mathrm{\partial }}_{x^k})\mathrm{.}$ Extend these coordinates off $S$ by keeping them constant along the $X_1$-flow, and use the flow parameter as $x^1$. Then $X_1={\mathrm{\partial }}_{x^1}$. Since each $Y_j$ commutes with $X_1$, its coefficients in these flow-box coordinates are constant along the $X_1$-flow, so the span identity on $S$ extends to $\mathrm{span}(Y_2,\dots ,Y_k)=\mathrm{span}({\mathrm{\partial }}_{x^2},\dots ,{\mathrm{\partial }}_{x^k})\mathrm{.}$ Therefore $\mathcal{D}=\mathrm{span}({\mathrm{\partial }}_{x^1},\dots ,{\mathrm{\partial }}_{x^k})$ on a neighborhood of $p$.

In those coordinates, the slices with [given] $x^{k+1},\dots ,x^n$ fixed are integral manifolds of $\mathcal{D}$. Thus the involutive case is integrable, proving 2 => 1.

Hence integrability and involutivity are equivalent, and in the involutive [given] case the distribution is locally flat in coordinates as stated.

In Frobenius coordinates $(x^1,\dots ,x^n)$, the distribution is spanned by ${\mathrm{\partial }}_{x^1},\dots ,{\mathrm{\partial }}_{x^k}$. Define $F(x^1,\dots ,x^n):=(x^{k+1},\dots ,x^n)\mathrm{.}$ Its differential has rank $n-k$, so $F$ is a submersion.

A tangent vector lies in $\ker dF$ exactly when its last $n-k$ coordinate components vanish, so precisely when it is a linear combination of ${\mathrm{\partial }}_{x^1},\dots ,{\mathrm{\partial }}_{x^k}$. Hence $\ker dF=\mathcal{D}$.

Therefore every involutive distribution is locally the common kernel of $n-k$ smooth first integrals.

Because $F$ is a submersion, $q$ is a regular value and the level set [given] $F^{-1}(q)$ is an embedded submanifold. Its tangent space at each point is the kernel of the differential of $F$. Therefore each connected component of $F^{-1}(q)$ is an integral manifold of $\ker dF$.

Repeating the same argument at every point of $M$ shows that each point [given] lies on such a connected component, so $\ker dF$ is integrable. Since an integral manifold of $\ker dF$ stays inside one level set of $F$, the maximal connected integral manifolds are exactly the connected components of the fibres.

Hence the kernel of a submersion is integrable with leaves equal to fibre [given] components.

Write $F=(f^1,\dots ,f^r)$. The tangency assumption says [given] $X(f^a)=0=Y(f^a)$ for every component $f^a$.

Therefore [given] $[X,Y](f^a)=X(Y(f^a))-Y(X(f^a))=0$ for every $a$. This means $dF([X,Y])=0$, so $[X,Y]$ is again tangent to the kernel family.

Thus kernel distributions are closed under brackets whenever they are [given] defined as smooth distributions.

The transverse coordinate map $y\circ i:C\to {\mathbb{R}}^{n-k}$ has zero [given] differential. Indeed, the tangent image of $i$ is $\mathcal{D}$, and in a flat chart the distribution is exactly the kernel of $dy$.

A smooth map with zero differential is locally constant, hence constant on [given] each connected component of its domain. Therefore $y\circ i$ is constant on $C$.

The image $i(C)$ is therefore contained in the slice with that fixed [given] transverse coordinate, namely in a single plaque of the flat chart.

Apply the previous lemma to the connected integral manifold $P_1$ inside a [given] flat chart producing $P_2$. Near $p$, the set $P_1$ must lie in the plaque of that chart through $p$, namely in $P_2$.

Reversing the roles of $P_1$ and $P_2$ gives the opposite inclusion on [given] possibly smaller neighborhoods. Intersecting those neighborhoods yields an open set $W$ with $P_1\cap W=P_2\cap W$.

Thus plaques through the same point determine the same germ. [given] ∎

Reflexivity holds because the constant curve at any point is piecewise [given] smooth and has derivative $0\in \mathcal{D}$.

Symmetry holds because reversing a tangent piecewise smooth curve negates [given] its derivative but keeps it inside the same linear subspaces.

Transitivity holds because concatenating two tangent piecewise smooth curves [given] produces another piecewise smooth curve with the same tangency property.

Therefore ${\sim }_{\mathcal{D}}$ is an equivalence relation. [given] ∎

By [L1], restrictions of flat charts to coordinate boxes form an open cover of $M$. By [L5], [L6], and [A2], choose a countable such cover $(U_m,{\phi }_m{)}_{m\in \mathbb{N}}$. A plaque through a point of $L_p$ lies in $L_p$: [L9] joins its points within the plaque, and its tangent spaces are $\mathcal{D}$. Hence the plaques of the chosen cover that meet $L_p$ cover $L_p$.

Let $i:N\to M$ be a connected integral manifold with $i(q)=p$, and put $S=i^{-1}(L_p)$. For $r\in N$, choose a flat-chart domain $U$ about $i(r)$ and let $C$ be the connected component of $i^{-1}(U)$ containing $r$. By [L7], $C$ is an open neighborhood of $r$, and [L3] maps it into one plaque. If $r\in S$, that plaque lies in $L_p$; if $r\mathrm{\notin }S$, it lies in a different reachability class. Thus $S$ and its complement are open. Since $N$ is connected and $q\in S$, one has $S=N$, so $i(N)\subseteq L_p$.

Fix one chosen-cover plaque $P_0$ through $p$. If a chosen-cover plaque $P$ lies in $L_p$, then for each $m$ the components of $P\cap U_m$, viewed in the intrinsic Euclidean-slice topology of $P$, form an at most countable family by [L7]. Applying [L3] to the plaque inclusion shows that each such component lies in one plaque of $U_m$, so $P$ meets at most countably many chosen-cover plaques. By [L8], the same is true across all $m$. Starting from $P_0$, take all neighbors at each finite stage and then the union over the countably many finite stages; [L8] makes the resulting family ${\mathcal{P}}_p$ at most countable. Every member lies in $L_p$ by [L9]. Conversely, on each smooth segment of a tangent path the transverse flat-chart coordinates have zero derivative and are constant by [L10], so the path is locally contained in a chosen-cover plaque. Compactness of its parameter interval gives a finite plaque subdivision from $P_0$ to a plaque containing its endpoint. Hence ${\mathcal{P}}_p$ covers $L_p$.

Give each $P\in {\mathcal{P}}_p$ its Euclidean slice coordinates. By [L2], overlapping plaque coordinates have smooth transition maps, so these patches form a smooth atlas. It is countable by step 2.1, and each Euclidean patch has a countable basis, so [L8] makes the induced topology second countable. The inclusion $j_p$ is continuous and injective in these patches; since $M$ is Hausdorff by [L5], distinct points of $L_p$ have disjoint inverse-image neighborhoods, so the induced topology is Hausdorff. It is locally Euclidean by construction. Therefore it is a smooth $k$-manifold, and in plaque coordinates $j_p$ is an injective immersion with tangent image $\mathcal{D}$.

The class $L_p$ is path connected in this topology: a tangent path from $p$ to any of its points admits the finite plaque subdivision used in step 2.1, and is continuous in each plaque chart. Hence $j_p:L_p\hookrightarrow M$ is a connected integral manifold.

The equation $j_p\circ \mathrm{\Phi }=i$ now forces a unique set map $\mathrm{\Phi }:N\to L_p$. For any flat-chart domain $U$, [L3] maps each connected component of $i^{-1}(U)$ into one plaque, and in that plaque chart $\mathrm{\Phi }$ has the same coordinate expression as $i$. Hence $\mathrm{\Phi }$ is smooth, and uniqueness follows from injectivity of $j_p$.

It remains to prove uniqueness of the smooth structure. If $k=0$, every plaque and hence the connected leaf $L_p$ is the singleton $\{p\}$, which has only its unique zero-manifold structure. Assume $k\ge 1$ and let another smooth structure on $L_p$ make $j_p$ a connected integral injective immersion. By [L3], every point has a connected neighborhood in that alternative structure whose image lies in one plaque of step 3.1. The coordinate expression of $j_p$ from that neighborhood to the plaque is between $k$-manifolds and has invertible derivative because both tangent images equal $\mathcal{D}$; [L4] makes it a local diffeomorphism. Thus the alternative charts and the plaque charts are smoothly compatible in both directions, so the structures coincide.

Steps 3.1 and 4.1 give the asserted connected integral leaf structure, steps 1.2 and 4.2 give the universal factorization, and step 4.3 proves uniqueness; consequently $L_p$ is the unique maximal connected integral manifold through $p$.

Every point of $M$ lies in its own equivalence class, so the union of the [given] maximal leaves is all of $M$.

Distinct equivalence classes are disjoint. Therefore distinct maximal [given] connected integral manifolds are disjoint.

Hence the maximal connected integral manifolds partition $M$. [given] ∎

In a regular foliation chart $(x,y)$, declare the tangent distribution to [given] be the span of ${\mathrm{\partial }}_{x^1},\dots ,{\mathrm{\partial }}_{x^k}$. Because overlap maps send plaque directions to plaque directions, these local $k$-planes patch to a smooth rank-$k$ distribution. Plaques are local integral manifolds, so the distribution is integrable.

Conversely, let $\mathcal{D}$ be an integrable rank-$k$ distribution. By the [given] local Frobenius theorem, every point has a flat chart $(x,y)$ for $\mathcal{D}$. Choose a covering by sufficiently small restrictions of these charts, refining overlap domains into plaque-coordinate neighborhoods. On each such overlap the new transverse coordinate has differential zero in every old plaque direction, so it is locally a function only of the old transverse coordinate. The refined charts therefore have transitions of the form required by a regular foliation atlas. This asserts existence of a compatible refined atlas; it does not claim that every unrestricted flat chart belongs to one common atlas.

In that atlas the plaques are precisely the local integral pieces of [given] $\mathcal{D}$, so the global leaves are exactly the maximal connected integral manifolds. The two constructions therefore recover one another.

Thus regular foliations and integrable distributions correspond. [given] ∎